Expectation values, the virial split and uncertainty

The definitions a = (X + iP)/√2 and a† = (X − iP)/√2 invert to X = (a + a†)/√2 and P = i(a† − a)/√2, and restoring the scales x₀ = √(ħ/mω) and p₀ = √(mħω) gives x = √(ħ/2mω)(a + a†) and p = i√(mħω/2)(a† − a). Every power of x or p is now a polynomial in a and a†, and the diagonal element ⟨n|⋅|n⟩ survives only on the terms that raise exactly as often as they lower. An odd power has no such term, so ⟨n|x|n⟩ = ⟨n|p|n⟩ = 0 on every rung. Parity would give the same two zeros, but the algebra gives strictly more: it also hands you the off-diagonal elements, ⟨m|x|n⟩ = √(ħ/2mω)(√n δ_{(m, n−1)} + √(n+1) δ_{(m, n+1)}), which is exactly the Δn = ±1 selection rule for an electric-dipole transition on a harmonic ladder.

x² = (ħ/2mω)(a + a†)² = (ħ/2mω)(a² + a†² + a a† + a† a). The first two terms move the ket two rungs and drop out of the diagonal element; the last two combine through a a† = a† a + 1 into 2N + 1. So ⟨n|x²|n⟩ = (ħ/2mω)(2n + 1) = (ħ/mω)(n + ½). The same move on p² = −(mħω/2)(a† − a)² leaves ⟨n|p²|n⟩ = mħω(n + ½), the sign flip supplied by the i². Two checks are worth doing. Recombining them gives ⟨p²⟩/2m + ½mω²⟨x²⟩ = (n + ½)ħω = Eₙ, as it must. And at n = 0 the position result reads ⟨x²⟩ = ħ/2mω, precisely the variance of the Gaussian |ψ₀|² ∝ exp(−mωx²/ħ) — the ladder has reproduced an integral you never performed.

Nothing in a stationary state changes, so d⟨xp⟩/dt = 0. Ehrenfest turns that into a commutator: d⟨xp⟩/dt = (i/ħ)⟨[H, xp]⟩. With [p², x] = −2iħp and [V(x), p] = iħ dV/dx, the commutator is [H, xp] = −iħ(p²/m − x dV/dx), so the condition reads 2⟨T⟩ = ⟨x dV/dx⟩. That is the quantum virial theorem, and note what it needs: a stationary state, not a time average. For a homogeneous potential V ∝ xᵏ it collapses to 2⟨T⟩ = k⟨V⟩. The oscillator has k = 2, hence ⟨T⟩ = ⟨V⟩ = Eₙ/2 = (n + ½)ħω/2 — which reproduces the two second moments of the previous section with no extra work. Hydrogen has k = −1 and therefore 2⟨T⟩ = −⟨V⟩: a positive kinetic share half the size of a negative potential one. Equal halves are a property of k = 2, not of bound states.

Because ⟨x⟩ = ⟨p⟩ = 0, the spreads are just the square roots of the second moments, so Δx Δp = √[(ħ/mω)(n + ½)] × √[mħω(n + ½)] = (n + ½)ħ. Mass and frequency cancel completely; only the rung number survives. The Robertson floor is ħ/2 for every state, and the Robertson–Schrödinger refinement adds a covariance term ½⟨{x, p}⟩ − ⟨x⟩⟨p⟩ that vanishes here: xp = (iħ/2)(1 − a² + a†²) gives ⟨n|xp|n⟩ = iħ/2, so ⟨{x, p}⟩ = 0 on every rung. The bound therefore stays at ħ/2 for all n, and only n = 0 attains it. The reason is the saturation condition itself — equality in Cauchy–Schwarz, with the covariance term zero, demands (p − ⟨p⟩)|ψ⟩ = iκ(x − ⟨x⟩)|ψ⟩ with real κ, a first-order differential equation in the position basis whose normalisable solutions are Gaussians. Of the Fock states only |0⟩ is Gaussian; |1⟩ already carries a node.

For CO, ħω = 269 meV and the reduced mass is μ = 1.139 × 10⁻²⁶ kg, so ω = 4.086 × 10¹⁴ rad s⁻¹ and the length scale is x₀ = √(ħ/μω) = 4.76 pm. In the ground state Δx = x₀/√2 = 3.37 pm and Δp = 1.57 × 10⁻²³ kg m s⁻¹, whose product is 5.27 × 10⁻³⁵ J s = ħ/2 to the digit. The energy split is ⟨T⟩ = ⟨V⟩ = 67.3 meV, half of the 134.5 meV zero-point energy. Against a 113 pm bond a 3.37 pm spread is 3.0%, small enough that a quadratic fit to the true internuclear potential is honest. Climb to n = 10 and Δx = x₀√10.5 = 15.4 pm, or 14% of the bond, where the cubic term dropped from the Taylor expansion is no longer negligible. One coincidence deserves naming: the classical turning point is Aₙ = √2 Δx, and a classical oscillator's time-averaged x² is also A²/2, so the quantum and classical second moments agree at every n even though the distributions do not.

Completeness gives the next moment in one line: ⟨n|x⁴|n⟩ = Σₘ |⟨m|x²|n⟩|², where only m = n − 2, n, n + 2 contribute, so with s = ħ/2mω the result is s²[(2n + 1)² + n(n − 1) + (n + 1)(n + 2)] = 3(2n² + 2n + 1)s². Perturb the Hamiltonian by λx⁴ and the first-order shift is 3(2n² + 2n + 1)λs², which makes the spacing Eₙ₊₁ − Eₙ = ħω + 12λs²(n + 1) — the uniform ladder is gone, and with it the evenly spaced spectrum that the commutator alone produced. The virial statement degrades in step: 2⟨T⟩ = ⟨mω²x² + 4λx⁴⟩ = 2⟨V₂⟩ + 4⟨V₄⟩, so subtracting ⟨V⟩ = ⟨V₂⟩ + ⟨V₄⟩ leaves the exact identity ⟨T⟩ − ⟨V⟩ = ⟨λx⁴⟩. Hellmann–Feynman fixes its size, since ∂E/∂λ = ⟨x⁴⟩, so at leading order the ground-state imbalance is 3λs². Equal shares were never a quantum principle; they were the k = 2 special case.